Medium
Given two strings s1
and s2
, return true
if s2
contains a permutation of s1
, or false
otherwise.
In other words, return true
if one of s1
’s permutations is the substring of s2
.
Example 1:
Input: s1 = “ab”, s2 = “eidbaooo”
Output: true
Explanation: s2 contains one permutation of s1 (“ba”).
Example 2:
Input: s1 = “ab”, s2 = “eidboaoo”
Output: false
Constraints:
1 <= s1.length, s2.length <= 104
s1
and s2
consist of lowercase English letters.public class Solution {
public boolean checkInclusion(String s1, String s2) {
int n = s1.length();
int m = s2.length();
if (n > m) {
return false;
}
int[] cntS1 = new int[26];
int[] cntS2 = new int[26];
for (int i = 0; i < n; i++) {
cntS1[s1.charAt(i) - 'a']++;
}
for (int i = 0; i < n; i++) {
cntS2[s2.charAt(i) - 'a']++;
}
if (check(cntS1, cntS2)) {
return true;
}
for (int i = n; i < m; i++) {
cntS2[s2.charAt(i - n) - 'a']--;
cntS2[s2.charAt(i) - 'a']++;
if (check(cntS1, cntS2)) {
return true;
}
}
return false;
}
private boolean check(int[] cnt1, int[] cnt2) {
for (int i = 0; i < 26; i++) {
if (cnt1[i] != cnt2[i]) {
return false;
}
}
return true;
}
}